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2 years ago

Which type of vehicle is used routinely at construction and mining sites?

Engineering

2 answers:

Semmy [17]2 years ago

6 0

Answer:

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

Explanation:

it really depends on the job.

satela [25.4K]2 years ago

5 0

We are required to state the type of vehicle which is routinely used at the construction site.

The vehicle which is routinely used at the construction site is called hydraulic vehicle.

Hydraulics as the name implies is pertaining to water. Hydraulic vehicle is a vehicle used on the construction site. The vehicle makes use of liquids to function. The fluid powers the engine of the vehicle do as to generate power.

Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago

Can someone help me with this maze shown below.

Gnoma [55]

We can’t see the maze

3 0

2 years ago

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

i (t) = 6*e^(-2*t)

v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

- The amount of charge Q delivered can be determined by:

dQ = i(t) . dt

$Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

$W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

$W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$

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3 years ago

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Jlenok [28]

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3 years ago

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Dafna11 [192]

Answer:

the 1st one i think

Explanation:

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3 years ago